Answer to Question #254389 in Calculus for Hasni

Question

y=sec^(−1) (x) ⇔ secy=x,f or y∈[0,π/2)∪[π,3π/2).

With this choice of principal branch the derivative of sec^(−1) (x) is

d/dx*sec^-1 (x)= 1/x*√(x^2 -1)

However, as mentioned in the side remark in Section 4.9, there is another common

definition of sec^−1(x), which is y=sec^(−1) (x), which is

y=sec^−1⁡(x) ⇔ sec⁡y=x, for y∈[0,π/2)∪(π/2,π].

Show that, with this choice of principal branch to define the inverse secant function, its derivative becomes

d/dx*sec^(−1) (x)=1/|x|*√(x^2 -1)

Answer 1

$secy=x$

$(secy)'=y'secytany=1$

$y'=\frac{1}{secytany}$

$tany=\sqrt{sec^2y-1}=\sqrt{x^2-1}$

$\frac{d}{dx}sec^{-1}x=\frac{1}{x\sqrt{x^2-1}}$